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Learning Math Home
Data Session 5: Solutions
 
Session 5 Part A Part B Part C Part D Part E Homework
 
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Solutions for Session 5, Part A

See solutions for Problems: A1 | A2 | A3 | A5 | A6| A7| A8| A9


Problem A1

The mean is 5. To calculate the mean, add the numbers in each stack. The sum is 45. Then divide by the total number of stacks, 9, to find the mean.

<< back to Problem A1


 

Problem A2

a. 

The mean is still 5.

b. 

Since the total number of coins remains 45, and the total number of stacks remains 9, the mean must still be 45 / 9 = 5.

c. 

You could increase or decrease the number of coins, or you could increase or decrease the number of stacks.

<< back to Problem A2


 

Problem A3

a. 

As before, the mean is 5.

b. 

Here, the mean is equal to the number of coins in each stack. Since all the stacks have the same number of coins, the average is equal to the number of coins in any given stack.

<< back to Problem A3


 

Problem A5

a. 

The median is not the fifth stack because the allocation is not in order. It must first be placed in order.

b. 

Here's the ordered allocation:

The stack in position (5), the middle, has 5 coins, so the median size is 5. For this allocation, the median is the same as the mean.

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Problem A6

Answers will vary. Here is one possible allocation:

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Problem A7

a. 

Answers will vary. Here is one such allocation, with a median size of 4 coins:

b. 

As before, the mean must remain 5, since the number of coins and the number of stacks had not changed.

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Problem A8

Answers will vary. Here is one such allocation, with a median size of 6 coins:

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Problem A9

The smallest possible value for the median is 1. It cannot be smaller, since each stack must contain at least 1 coin. One possible allocation with this median is {1, 1, 1, 1, 1, 10, 10, 10, 10}; many other allocations are possible if a stack may contain more than 10 coins.

The largest possible value for the median is 8. One way to think about this problem is to recognize that there must be 4 stacks below the median, and each must have at least 1 coin. This leaves 41 or fewer coins for the remaining five stacks, and the median must be the smallest of these stacks. Using an equal-shares allocation of the remaining 41 coins results in 4 stacks of 8 coins and 1 of 9, so the median cannot be greater than 8. There are only two possible allocations with this median: {1, 1, 1, 1, 8, 8, 8, 8, 9} and {1, 1, 1, 2, 8, 8, 8, 8, 8}.

<< back to Problem A9


 

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